1) 0.89% m/v = 0.89 grams of NaCl / 100 ml of solution
=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution
2) Molarity = M = number of moles of solute / liters of solution
=> calculate the number of moles of 8.9 grams of NaCl
3) molar mass of NaCl = 23.0 g /mol + 35.5 g/mol = 58.5 g / mol
4) number of moles of NaCl = mass / molar mass = 8.9 g / 58.5 g / mol = 0.152 mol
5) M = 0.152 mol NaCl / 1 liter solution = 0.152 M
Answer: 0.152 M
The mass of the product is <em>98.78 g.</em>
The word equation is
aluminum + chlorine → product
20.00 g + 98.78 g → <em>x</em> g
If each reactant is completely consumed, the <em>Law of conservation of Mass </em>tells us the mass of the product must be 98.78 g.
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Hello!
The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.
We have the following data:
mo (initial mass) = 53.3 mg
m (final mass after time T) = ? (in mg)
x (number of periods elapsed) = ?
P (Half-life) = 10.0 minutes
T (Elapsed time for sample reduction) = 25.9 minutes
Let's find the number of periods elapsed (x), let us see:
Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:
I Hope this helps, greetings ... DexteR! =)
Answer:
pH = 8.34
Explanation:
The equilbriums of the amphoteric HCO₃⁻ (Ion of NaHCO₃) are:
H₂CO₃ ⇄ <em>HCO₃⁻</em> + H⁺ Ka1 <em>-Here, HCO₃⁻ is acting as a base-</em>
<em>HCO₃⁻</em>⇄ CO₃²⁻ + H⁺ Ka2 <em>-Here, is acting as an acid-</em>
Where Ka1 = 4.3x10⁻⁷ and Ka2 = 4.8x10⁻¹¹. As pKa = -log Ka:
pKa1 = 6.37; pKa2 = 10.32
As the pH of amphoteric salts is:
pH = (pKa1 + pKa2) / 2
<h2>pH = 8.34</h2>
